Quadratic Lessons Assignment
By Mohamed B.
Introduction
Expanding
I have a variable and a number in brackets multiplied by another variable and a number in brackets.
(x-12) (x+3)
Now I am going to use distributive property to put it in standard form.
x times x is X2
x times 3 is 3x
-12 times x is -12x
-12 times 3 is -36
If you put it all together, you will get X2 - 9x -36
If you are still stuck, look at the image below. It is another example.
Factoring
Simple Trinomial
I have a equation like this x2 + 5x + 6
So the question is what 2 numbers multiply's to 6 and add to 5. The answer is 3 and 2. This is because 2 times 3 is 6 and 2 plus 3 is 5. Easy right?
Now we have to put it in the expanded form which is like this:
(x+3)(x+2) -- if you expand this you will get the same equation as the one above. It is a cycle!
Decomposition
Now that was the simplest form. There are still 2 more forms to cover. The next one is called decomposition.
If I have the same equation as the one above but instead of nothing in front of the A value, I have a number. This is then when we use decomposition.
So lets say I have this :
2x2 + 5x + 3 - I cant factor this because there is nothing in common. Therefore, I will multiply the A value by the C value and do the same thing as before; which is to find 2 numbers that multiply to the (new value) and add to the B value.
ok so 2 times 3 is 6. What multiply's to 6 and adds to 5. 3 and 2. Now we put it like this:
2x2 +3x +2x +3 - what you do is that you remove the b value and put the 2 new numbers in and add an x beside it.
Now we have to factor by grouping. This is a sub category. Basically you take the first 2 numbers/variables and you factor and then you take the next two varialbes/numbers and you factor.
3/2x(4/3x+2)+3/2(4/3x+2)
Then if you join the things that are not in brackets together and take out 1 of the same things in brackets, you will get this:
(3/2x+2)(4/3x+2)
And there you go. By the way sorry, I had to use fractions. Usually fractions/decimals are not used in decomposition.
Difference Of Squares
The last factoring thing is called difference of squares.
Ok so lets say I have an equation like this
x2 - 49y2
If all these can be square rooted and there is a negative sign between them, we can find the difference of squares.
This is what you do.
( x+7y) (x-7y)
We get the root of it and then we do x + whatever the number and x - whatever the number. Also it can have more than 1 variable, different variable, etc. (not necessarily x.)
Still having trouble with difference of squares? Watch This !!
Completing The Square
1. If there is a A value factor it out. In this case their is
4(x2+0.5x + 1.5)
2. Take the C value out but multiply it by the factored A
4(x2+0.5x)+6
3. Divide the B value by 2 and square it. Then add the answer in the equation.
4(x2+0.5x + 0.0625)+6
4. The same number that you divided by 2 and squared, make it a negative and multiply it by the A. Then place it outside the bracket
4(x2+0.5x + 0.0625)+6 - 0.25 ( subtract 6 - .25)
= 4(x2+0.5x + 0.0625)+5.75
5. Now see that value that was divided and squared and added in the brackets. Get rid of it.
= 4(x2+0.5x)+5.75
6. And the last step is to divide the 0.5 or B value by 2 and square the whole bracket. Also get rid of the square on the x.
4(x+0.25)^2+5.75
And there you go. You have completed the square successfully!
Still stuck, this video should help.
Solving a Quadratic Equation
1rst way
Standard form
x^2 -2x-35 = 0
Now what 2 numbers multiply to -35 and add to -2. -7 and 5 right?
(x-7)(x+5) = 0
Now we have to solve for x. So what 2 numbers make 0. The first one is x = 7 and the second one is -5. If you put those variables in for the equation it will be 0. Here is how.
(7-7)(-5+5) = 0 ---- this equation makes sense. So the final answer is x = 7 and x = -5.
The 2nd way of solving is say I end up with something like this.
(2x-5)(2x+3) = 0
I have to solve for x right. So I will move the number to the other side and divide by the variable. You will end up with this.
x = 5/2 and x = -3/2
Still having trouble. This last video will help.